Question 9.13 (Sixth Edition)


Moderators: Chem_Mod, Chem_Admin

Steve Magana 2I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

Question 9.13 (Sixth Edition)

Postby Steve Magana 2I » Wed Feb 20, 2019 2:07 pm

Question: During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 degrees Celsius was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 degrees Celsius. Assume ideal behavior. What is the change in entropy of the gas?

I'm having trouble on this question, can someone help me out please? Thank you!

Kristen Kim 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:16 am

Re: Question 9.13 (Sixth Edition)

Postby Kristen Kim 2K » Wed Feb 20, 2019 2:51 pm

For this problem, there are two variables that are changing: volume and temperature. In order to find total entropy, you need to solve for two equations and add the two together to get the total value. We can use delta S = nRln(V2/V1) for the change in volume and delta S = nRln(T2/T1) for the change in temperature.

Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

Re: Question 9.13 (Sixth Edition)

Postby Stevin1H » Thu Feb 21, 2019 10:14 am

Since Entropy is a state function and we see that there are two variables changing (the volume and temperature), we can set up two different equations and add the total to find the total change in entropy. We have the equation delta S = nRln(V2/V1) for when the volume is changing and delta S = nRln(T2/T1) to account for a change in temperature. And after plugging in the values into both equations and adding them up, you should get -14.6 J/K.

Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

Re: Question 9.13 (Sixth Edition)

Postby 005199302 » Thu Mar 07, 2019 6:54 pm

Isn't the formula for entropy of temperature using C, not R? Is this a mistake in the book?

Return to “Entropy Changes Due to Changes in Volume and Temperature”

Who is online

Users browsing this forum: No registered users and 1 guest